[alogo] Reflexions of a line

Start with a given point D on side BC of an accute-angled triangle ABC and a direction DE. Reflect line DE on side AC to create line EF and reflect again EF on side AB to create line ED'. Find the direction of DE for which D'=D.

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]
[2_0] [2_1] [2_2]

The figure suggests the solution. Reflect D on sides {AC,AB} to obtain points {D2, D1}. Line D1D2 intersects these sides at points {E,F} and line DE has the required property.

Remark-1 The two orthogonals {EI,FI} to sides {AC,AB} and line AD pass through the same point I, which is the incenter of triangle DEF.
Remark-2 If we require that also at point D the lines {FD,DE} are reflexions of each other on side BC, then AD must be orthogonal to BC and triangle DEF coincides with the orthic triangle D0E0F0 of ABC.
Remark-3 Letting D take the positions on BC which are the reflexions of H (see its definition in [4]) on lines {AC,AB} we see that line D1D2 obtains the positions of lines {HK,HL}, which are the reflexions of BC on sides {AC,AB}.

Remark-4 For D varying on line BC, the corresponding lines D1D2 envelope the parabola (c) with focus at A and vertex at point J, which is the projection of A on side E0F0 of the orthic triangle. This identifies the E0F0 with the tangent to the parabola at its vertex and shows that the axis of the parabola is the isogonal conjugate of the altitude from A.

To prove this, take the reflexions {HB,HC} of line BC with respect to sides {AB,AC}. They intersect at a point H and line AH is easily proved to be orthogonal to side E0F0 of the orthic. Consider the reflexion A* of point A on line D1D2. Triangle D1A*D2 is equal to D1AD2 and remains all the time similar to itself as D changes position on D. Thus (see Similarly_Rotating.html ) A* moves on a line which is the directrix of the claimed parabola with focus at A.

Remark-5 Another proof of the last remark results by drawing parallelogram D1HD2M and showing that M moves on the line parallel to side E0F0 of the orthic at double distance from H. This identifies the parabola with the Artzt parabola of triangle HKL with respect to side KL.
Remark-6 Quadrangle HD1AD2 is cyclic. Its circumcircle is the circumcircle of triangle D1D2H, which has its sides tangent to the parabola (see remark-3). This circle passes through the focus A of the parabola as is always the case for triangles with sides tangent to a parabola (see [10] in ParabolaChords.html ).

The results on the envelope can be generalized for a segment of fixed length gliding on side BC. This is discussed in ReflectingSegmentEnds.html .

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